Let 
\[R\] be the region enclosed by the polar curve 
\[r=1+\theta\] to the left of the 
\[y\]-axis, as shown in the graph.
A polar curve is graphed. The x axis goes from negative 4 to 4. The curve is open and spiral shaped, starting at (1, 0) and moving counterclockwise through the following points. (1.25, 1), (0, 2.5), (negative 1.5, 3), (negative 4.1, 0), (negative 2, negative 5), (0, negative 5.8), and (4, negative 5). The area enclosed by the curve and axes in the second and third quadrants, between x = 0 and x = negative 4.1 and y = negative 5.8 and y = 3, is shaded, representing region R. All values estimated.
\[y\]
\[x\]
\[R\]
\[\small 1\]
\[\small 1\]
What is the area of 
\[R\]?
Use a graphing calculator and round your answer to three decimal places.


To find the area of region \(R\) enclosed by the polar curve \(r = 1 + \theta\), we use the following formula for the area enclosed by a polar curve between two angles \(\theta_1\) and \(\theta_2\):

\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta
\]

For the given curve \(r = 1 + \theta\), the area of region \(R\) is to the left of the \(y\)-axis. This implies that we want the area corresponding to \(\theta\) values from \( \theta = \frac{\pi}{2} \) (along the positive \(y\)-axis) to \( \theta = \frac{3\pi}{2} \) (along the negative \(y\)-axis).

We can now substitute \(r = 1 + \theta\) into the formula and evaluate the integral for \(\theta\) ranging from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\):

\[
A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (1 + \theta)^2 \, d\theta
\]

### Steps:

1. Expand \((1 + \theta)^2\):
\[
(1 + \theta)^2 = 1 + 2\theta + \theta^2
\]

2. Substitute into the integral:
\[
A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (1 + 2\theta + \theta^2) \, d\theta
\]

3. Break the integral into three parts:
\[
A = \frac{1}{2} \left[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 1 \, d\theta + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2\theta \, d\theta + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \theta^2 \, d\theta \right]
\]

Now, let's calculate each part of the integral.The area of region \(R\) enclosed by the polar curve \(r = 1 + \theta\), to the left of the \(y\)-axis, is approximately \(28.235\) square units.

Would you like more details on the steps or have any questions?

Here are some related questions to explore:

1. How would the area change if the polar curve was \(r = 2 + \theta\)?
2. How is the polar area formula derived, and why is it different from the Cartesian area formula?
3. Can this integral be solved geometrically without calculus?
4. What would the graph of \(r = 1 + 2\theta\) look like?
5. How does the curve behave as \(\theta\) approaches large positive or negative values?

**Tip:** When dealing with polar areas, always check the limits of \(\theta\) carefully, as they define which portion of the curve you're analyzing.

Let R be the region enclosed by the polar curve r = 1 + θ to the left of the y-axis. A polar curve is graphed. The x axis goes from negative 4 to 4. The curve is open and spiral shaped, starting at (1, 0) and moving counterclockwise through the following points: (1.25, 1), (0, 2.5), (negative 1.5, 3), (negative 4.1, 0), (negative 2, negative 5), (0, negative 5.8), and (4, negative 5). The area enclosed by the curve and axes in the second and third quadrants, between x = 0 and x = negative 4.1 and y = negative 5.8 and y = 3, is shaded, representing region R. What is the area of R? Use a graphing calculator and round your answer to three decimal places.

Learn how to calculate the area of a region enclosed by the polar curve r = 1 + θ to the left of the y-axis. This problem involves polar coordinates, integral calculus, and the polar area theorem. Suitable for students in Grades 10-12, this example demonstrates key mathematical techniques for finding areas using integrals in polar form.

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